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Let a non-zero column matrix $A_{m\times 1}$ be multiplied with a non-zero row matrix $B_{1\times n}$ to get a matrix $X_{m\times n}$ . Then how to find rank of $X$?

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2 Answers 2

Hint: To find the rank of a matrix, one usually row (or column) reduces it. In your case, all the rows (or columns) are multiples of one row (column). What can you conclude about the reduced matrix?

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Can't get it. I have just read basic meaning of rank and nothing more than that about rank. –  Mr.ØØ7 Apr 4 '13 at 11:11
    
@exploringnet: in case any two rows are linearly dependent (they are both multiples of $B$), can it happen that the rank is $2$? –  Ilya Apr 4 '13 at 11:30
    
@Ilya Yes , but then answer is $min(m,n)$ yes ? . I don't have answer with me. –  Mr.ØØ7 Apr 4 '13 at 11:35
    
@exploringnet: well, the rank of $\Bbb R^{m\times n}$ matrix is at most $\min(m,n)$. In your case, clearly, the matrix is very degenerate. –  Ilya Apr 4 '13 at 11:39
    
@exploringnet: the rank of a matrix is the dimension of the space its columns or rows span. Since all the rows of the product are multiples of the row matrix, what is the dimension of the space they span? Since all the columns of the product are multiples of the column matrix, what is the dimension of the space they span? –  robjohn Apr 4 '13 at 11:46

Hint: You always have

$$rk(AB)\leq min\{rk(A),rk(B)\}$$

It would be helpful to convince yourself of this fact. @DennisGulko's answer gives you an idea how.

Now can the rank of $AB$ be zero? That would mean that the matrix has only zero-entries...

Edit: I assume the entries of the matrix are elements of a field, i.e there are no zero-divisors.

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